Rotational Mechanics

Q1 [12]

Setup: Cylinder $I_{cm} = \tfrac12 MR^2$. Spool radius $r$, thread from top, force $F$ horizontal.

(a) [4]

• Forces: applied $F$ (horizontal, at height $R+r$... but torque about cm uses moment arm $r$), weight $Mg$, normal $N$, static friction $f$ at contact. [1]
• Translational: $F + f = Ma$ (taking $f$ positive in direction of $F$; sign found later). [1½]
• Rotational about cm: torque from $F$ = $F\cdot r$ (thread at radius $r$ from axis), torque from friction = $f\cdot R$. $F r + f R = I_{cm}\alpha = \tfrac12 MR^2 \alpha.$ [1½] (Direction convention: both $F$ and $f$ taken to produce forward rolling; signs resolve algebraically.)

(b) [5] Rolling: $a = R\alpha \Rightarrow \alpha = a/R$. [1] From translation: $f = Ma - F$. [1] Substitute into rotation: $Fr + (Ma - F)R = \tfrac12 MR^2 \cdot \tfrac{a}{R} = \tfrac12 MRa.$ [1] $Fr + MaR - FR = \tfrac12 MRa \Rightarrow F(r - R) = \tfrac12 MRa - MRa = -\tfrac12 MRa.$ $F(R - r) = \tfrac12 MRa \Rightarrow a = \frac{2F(R-r)}{MR}.$ [2]

Note: the sign/geometry of thread leaving the top gives the effective lever difference $(R-r)$. Clean result: $\boxed{a = \frac{2F(R-r)}{MR}}.$

(c) [3] $f = Ma - F = M\cdot\frac{2F(R-r)}{MR} - F = \frac{2F(R-r)}{R} - F = F\left(\frac{2R - 2r - R}{R}\right) = F\frac{(R-2r)}{R}.$ [2] $\boxed{f = \frac{F(R-2r)}{R}}$ Friction vanishes when $r = R/2$; for $r<R/2$ friction acts forward, for $r>R/2$ backward. [1]

Q2 [14]

(a) [5] For a sphere rolling down incline with $I = kMR^2$: $a = \frac{g\sin\theta}{1+k}.$ [2] Solid sphere $k = 2/5$: $a_s = \dfrac{g\sin\theta}{7/5} = \dfrac{5}{7}g\sin\theta.$ [1½] Hollow sphere $k = 2/3$: $a_h = \dfrac{g\sin\theta}{5/3} = \dfrac{3}{5}g\sin\theta.$ [1½]

(b) [4] From $L = \tfrac12 a t^2$, $t = \sqrt{2L/a}$, so $\frac{t_s}{t_h} = \sqrt{\frac{a_h}{a_s}} = \sqrt{\frac{3/5}{5/7}} = \sqrt{\frac{21}{25}} = \frac{\sqrt{21}}{5} \approx 0.9165.$ [3+1]

(c) [3] Solid sphere reaches first (smaller $k$ ⇒ larger acceleration ⇒ less energy in rotation). [1] Speed: $v = \sqrt{2aL}$, so $\frac{v_s}{v_h} = \sqrt{\frac{a_s}{a_h}} = \sqrt{\frac{5/7}{3/5}} = \sqrt{\frac{25}{21}} \approx 1.091.$ [2]

(d) [2] On the frictionless patch, no torque ⇒ $\omega$ stays constant; no horizontal force ⇒ $v$ stays constant. The sphere continues at the speed and spin it had; rolling condition $v=R\omega$ is still satisfied so it simply glides/rolls at constant velocity (no acceleration). [2]

Q3 [12]

(a) [3]

• Linear momentum: conserved — no external horizontal force on table (frictionless), impulsive contact is internal to the system. [1]
• Angular momentum about any fixed point: conserved — no external torque during impact. [1]
• Mechanical (kinetic) energy: NOT conserved — inelastic (sticking) collision. [1]

(b) [6] Place rod centre at origin; putty hits at $x = L/2$. CM of combined system from rod centre: $x_{cm} = \frac{m(L/2)}{M+m}.$ [1] Conserve angular momentum about the combined CM. Initial: only putty has momentum $mv_0$ at perpendicular distance $d = \tfrac{L}{2} - x_{cm}$ from CM: $L_i = m v_0\left(\frac{L}{2} - x_{cm}\right) = m v_0 \cdot \frac{L}{2}\cdot\frac{M}{M+m}.$ [2] Moment of inertia about combined CM: $I = \underbrace{\tfrac{1}{12}ML^2 + M x_{cm}^2}_{\text{rod (parallel axis)}} + m\left(\tfrac{L}{2}-x_{cm}\right)^2.$ [1½] Then $L_i = I\omega$: $\boxed{\omega = \frac{m v_0 \frac{ML}{2(M+m)}}{\frac{1}{12}ML^2 + M x_{cm}^2 + m(\frac{L}{2}-x_{cm})^2}}.$ [1½]

Simplified form (standard result): using $I = \frac{1}{12}ML^2 + \frac{mM}{m+M}\cdot\frac{L^2}{4}$, $\omega = \frac{m v_0 \frac{ML}{2(M+m)}}{\frac{1}{12}ML^2 + \frac{mM L^2}{4(m+M)}}.$

(c) [3] For $m = M/2$: let $M=2$, $m=1$, $L=1$, $v_0=1$ (ratios only). $x_{cm} = \frac{1\cdot0.5}{3} = 1/6$. $d = 0.5 - 1/6 = 1/3$. $I = \frac{1}{12}(2)(1) + 2(1/6)^2 + 1(1/3)^2 = \frac{1}{6} + \frac{2}{36} + \frac{1}{9} = \frac{1}{6}+\frac{1}{18}+\frac{2}{18}=\frac{3}{18}+\frac{3}{18}=\frac{6}{18}=\frac13.$ $L_i = 1\cdot1\cdot\frac13 = 1/3$. $\omega = L_i/I = (1/3)/(1/3) = 1$. KE final $= \tfrac12 I\omega^2 = \tfrac12(1/3)(1) = 1/6.$ KE initial $= \tfrac12 m v_0^2 = 1/2.$ Fraction lost $= 1 - \frac{1/6}{1/2} = 1 - \frac13 = \boxed{2/3}$. [3]

Q4 [10]

(a) [3] Gravity exerts a torque $\tau = mg\ell\sin\phi$ about the pivot, perpendicular to the spin angular momentum $L$. This torque changes the direction of $L$ (not magnitude), causing the axis to sweep around vertically — precession. Because $\vec\tau \perp \vec L$, $L$ rotates horizontally instead of the top toppling. [3]

(b) [4] $I = \tfrac12 mR^2 = \tfrac12(0.20)(0.04)^2 = \tfrac12(0.20)(0.0016) = 1.6\times10^{-4}\ \text{kg·m}^2.$ [1] $L = I\omega = 1.6\times10^{-4}\times150 = 0.024\ \text{kg·m}^2/\text{s}.$ [1] $\Omega = \frac{\tau}{L} = \frac{mg\ell}{I\omega} = \frac{0.20\times9.8\times0.03}{0.024} = \frac{0.0588}{0.024} = 2.45\ \text{rad/s}.$ [2]

(c) [3] $\Omega \propto 1/\omega$, so doubling spin halves precession: $\Omega' = 1.225\ \text{rad/s}$. [2] The formula assumes fast-top approximation $\Omega \ll \omega$; here $\Omega/\omega \approx 2.45/150 \approx 0.016 \ll 1$, so approximation is well justified. [1]

Q5 [12]

(a) [3] With smooth wall, wall exerts only horizontal normal $N_w$. Floor exerts vertical $N_f$ and horizontal friction $f$. Conditions:

• $\sum F_x = 0:\ N_w = f$ [1]
• $\sum F_y = 0:\ N_f = (M+m)g$ [1]
• $\sum \tau = 0$ (about base) [1]

(b) [6] $N_f = (20+60)(9.8) = 80\times9.8 = 784\ \text{N}.$ [1] Torque about base (take $L=1$ for ratios; weights at horizontal lever = position$\times\cos\theta$, wall force at $L\sin\theta$): Rod weight at $L/2$, person at $3L/4$. Balance: $N_w (L\sin\theta) = Mg\left(\tfrac{L}{2}\cos\theta\right) + mg\left(\tfrac{3L}{4}\cos\theta\right).$ [2] $N_w = \frac{\cos\theta}{\sin\theta}\left(\tfrac12 Mg + \tfrac34 mg\right)=\cot 60^\circ\left(\tfrac12(20)(9.8)+\tfrac34(60)(9.8)\right).$ $\tfrac12 Mg = 98$, $\tfrac34 mg = 441$; sum $=539\ \text{N}$. $\cot60° = 0.5774$. $N_w = 0.5774\times539 = 311.2\ \text{N}.$ [2] $f = N_w = 311.2\ \text{N}.$ [1]

(c) [3] $\mu_{min} = f/N_f = 311.2/784 = 0.397.$ [3] $\boxed{\mu_{min}\approx0.40}$

[
  {"claim":"Q2b time ratio ts/th = sqrt(21)/5 ≈ 0.9165","code":"a_s=Rational(5,7); a_h=Rational(3,5); ratio=sqrt(a_h/a_s); result=abs(float(ratio)-0.91652)<1e-3"},
  {"claim":"Q2c speed ratio vs/vh = sqrt(25/21) ≈ 1.0911","code":"a_s=Rational(5,7); a_h=Rational(3,5); ratio=sqrt(a_s/a_h); result=abs(float(ratio)-1.09109)<1e-3"},
  {"claim":"Q3c fraction of KE lost = 2/3 for m=M/2","code":"M=2; m=1; L=1; v0=1; xcm=Rational(m*L,2)/(M+m); d=Rational(L,2)-xcm; I=Rational(1,12)*M*L**2+M*xcm**2+m*d**2; Li=m*v0*d; w=Li/I; KEf=Rational(1,2)*I*w**2; KEi=Rational(1,2)*m*v0**2; frac=1-KEf/KEi; result=frac==Rational(2,3)"},
  {"claim":"Q4b precession Omega = 2.45 rad/s","code":"m=0.20; g=9.8; l=0.03; R=0.04; w=150; I=0.5*m*R**2; Om=(m*g*l)/(I*w); result=abs(Om-2.45)<0.02"},
  {"claim":"Q5c mu_min ≈ 0.397","code":"import math; Nf=80*9.8; Nw=(1/math.tan(math.radians(60)))*(0.5*20*9.8+0.75*60*9.8); mu=Nw/Nf; result=abs(mu-0.397)<0.005"}
]